Question 8 of 23intermediate🔧 ApplyNumerical3 marks

A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s−1^{-1}? (g=9.8 m s−2g = 9.8 \text{ m s}^{-2})

Correct Answer

The total time is the sum of the time taken by the stone to fall and the time taken by the sound to travel up.

Step 1: Time for stone to fall (t1t_1) Using the equation of motion h=ut+12gt2h = ut + \frac{1}{2}gt^2 with initial velocity u=0u=0, height h=300h=300 m, and g=9.8g=9.8 m s−2^{-2}:

300=12(9.8)t12300 = \frac{1}{2}(9.8)t_1^2 t12=6009.8=61.22t_1^2 = \frac{600}{9.8} = 61.22

t1=61.22≈7.82t_1 = \sqrt{61.22} \approx 7.82 s.

Step 2: Time for sound to travel (t2t_2) The speed of sound is v=340v = 340 m s−1^{-1}. The sound travels the same height h=300h=300 m. t2=distancespeed=300340≈0.88t_2 = \frac{\text{distance}}{\text{speed}} = \frac{300}{340} \approx 0.88 s.

Step 3: Total Time Total time T=t1+t2=7.82+0.88=8.70T = t_1 + t_2 = 7.82 + 0.88 = 8.70 s. The splash is heard at the top after 8.70 s.

Exercise: EXERCISES | Q: 14.2 | (Page 20)
For More Understanding

Explanation

The problem is solved in two stages: the downward motion of the stone under gravity and the upward propagation of sound. The context explicitly states that the speed of a mechanical wave depends only on the properties of the medium and not on the velocity of the source, which justifies using the constant speed of sound (340 m/s) for the return journey. The time of fall is calculated using kinematic equations for uniformly accelerated motion, and the sound travel time is calculated using the basic wave equation time=distance/speedtime = distance/speed.

Solution Steps

  1. Step 1: Calculate the time of fall (t1t_1) using h=12gt12h = \frac{1}{2}gt_1^2. Substituting h=300h=300 m and g=9.8g=9.8 m s−2^{-2}, we get t1=6009.8≈7.82t_1 = \sqrt{\frac{600}{9.8}} \approx 7.82 s.

  2. Step 2: Calculate the time for sound to travel (t2t_2) using t2=hvt_2 = \frac{h}{v}. Substituting h=300h=300 m and v=340v=340 m s−1^{-1}, we get t2=300340≈0.88t_2 = \frac{300}{340} \approx 0.88 s.

  3. Step 3: Add the two times to find the total duration: T=7.82+0.88=8.70T = 7.82 + 0.88 = 8.70 s.

PreviousNext